Stability and convergence of a conservative finite difference scheme for the modified Hunter–Saxton equation

Abstract

The modified Hunter–Saxton equation models the propagation of short capillary-gravity waves. As the equation involves a mixed derivative, its initial value problem on the periodic domain is much more complicated than the standard evolutionary equations. Although its local well-posedness has recently been proved, the behavior of its solution is yet to be investigated. In this paper, a conservative finite difference scheme is derived as a reliable numerical method for this problem. Then, the stability of the numerical solution in the sense of the uniform norm, and the uniform convergence of the numerical solutions to sufficiently smooth exact solutions are rigorously proved. Discrete conservation laws are used to overcome the difficulty due to the mixed derivative.

Appendices

Appendix

A Relation to the method of Miyatake et al. [18]

Miyatake et al. [18] derived the initial value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{t} = \left( \partial _x^{\dagger } \right) ^2 \left( -\frac{1}{2} \left( u^2 \right) _{xxx} + 2 \omega u_x + \frac{1}{2} \left( u_x^2 \right) _x \right) \qquad &{} (t \in (0,T), x \in \mathbb {S}), \\ u (0,x) = u_0 (x) &{} (x \in \mathbb {S}), \end{array}\right. } \end{aligned}$$

(A.1)

by assuming that the underdetermined form (1.5) has a traveling wave solution ($\partial _x^{\dagger }$ is a pseudo-inverse of the differential operator; see [18] for details). Accordingly, they devised underdetermined discretization (4.3) for preservation of ${\mathscr {H}}$ and employed the Moore–Penrose inverse $\sigma ^{\langle 2 \rangle }_x$ of the second-order central difference $\delta ^{\langle 2 \rangle }_x$. Thus, their scheme can be written in the form

$$\begin{aligned} {\left\{ \begin{array}{ll} \delta ^+_t u^{\left( m \right) }_{ k } = \sigma ^{\langle 2 \rangle }_x \left( {\mathscr {A}}_{\mathrm {d}} \big ( u^{\left( m+1/2 \right) }_{ } \big ) u^{\left( m+1/2 \right) }_{ k } \right) \quad &{} (m=0,1,\dots ,M-1;k \in \mathbb {Z}),\\ u^{\left( 0 \right) }_{ k } = u_0 (k \varDelta x) &{} (k \in \mathbb {Z}),\\ u^{\left( m \right) }_{ k+K } = u^{\left( m \right) }_{ k } &{} ( m = 0,1,\dots ,M-1, k \in \mathbb {Z} ). \end{array}\right. } \end{aligned}$$

(A.2)

Although they were interested in traveling wave solutions, this scheme can be employed for the problem (A.1) even when the solution is not actually a traveling wave.

Actually, initial value problems (4.7) and (A.1) are equivalent for sufficiently smooth solutions. Thus, both our scheme (4.6) and that of Miyatake, Cohen, Furihata, and Matsuo [18] (A.2) can be regarded as a numerical scheme for the artificial initial value problem (4.7). Moreover, although the motivations and derivations of these schemes are different, the resulting schemes (4.6) and (A.2) happen to be equivalent for the generalized problem (4.7) (or (A.1)) as follows.

Theorem A.1

If $u^{\left( m \right) }_{ k }$ is a solution of (4.6), then it is also a solution of (A.2), and vice versa.

Proof

To prove the former part, we assume that $u^{\left( m \right) }_{ k }$ is a solution of (4.6). Then, owing to the conservation of $\sum _{k=1}^K u^{\left( m \right) }_{ k } \varDelta x$, we observe that $\delta ^+_t u^{\left( m \right) }_{ } \in \mathop {\mathrm {car}}\nolimits \big ( \delta ^{\langle 2 \rangle }_x \big )$ holds. Therefore, on the basis of the property of the Moore–Penrose inverse, it holds that

$$\begin{aligned} \delta ^+_t u^{\left( m \right) }_{ k } = \sigma ^{\langle 2 \rangle }_x \left( {\mathscr {A}}_{\mathrm {d}} \big ( u^{\left( m+1/2 \right) }_{ } \big ) u^{\left( m+1/2 \right) }_{ k } \right) , \end{aligned}$$

which proves the former part.

On the other hand, for the latter part, we assume that $u^{\left( m \right) }_{ k }$ is a solution of (A.2). By operating with $\delta ^{\langle 2 \rangle }_x$ on both sides of the scheme in (A.2), we see that

$$\begin{aligned} \delta ^{\langle 2 \rangle }_x \delta ^+_t u^{\left( m \right) }_{ k } = \delta ^{\langle 2 \rangle }_x \sigma ^{\langle 2 \rangle }_x \left( {\mathscr {A}}_{\mathrm {d}} \big ( u^{\left( m+1/2 \right) }_{ } \big ) u^{\left( m+1/2 \right) }_{ k } \right) . \end{aligned}$$

Since $\delta ^{\langle 2 \rangle }_x \sigma ^{\langle 2 \rangle }_x = P$, it is sufficient to prove that ${\mathscr {A}}_{\mathrm {d}} \big ( u^{\left( m+1/2 \right) }_{ } \big ) u^{\left( m+1/2 \right) }_{ k } \in \mathop {\mathrm {range}}\nolimits \big ( \delta ^{\langle 2 \rangle }_x \big )$. Therefore,

$$\begin{aligned} \sum _{k=1}^K {\mathscr {A}}_{\mathrm {d}} \left( u^{\left( m+1/2 \right) }_{ } \right) u^{\left( m+1/2 \right) }_{ k } \varDelta x&= \sum _{k=1}^K \delta ^{\langle 1 \rangle }_x \left( \left( 2 \omega - \left( \delta ^{\langle 2 \rangle }_x u^{\left( m+1/2 \right) }_{ k } \right) \right) u^{\left( m+1/2 \right) }_{ k } \right) \varDelta x \\&\quad - \sum _{k=1}^K \left( \delta ^{\langle 2 \rangle }_x u^{\left( m+1/2 \right) }_{ k } \right) \left( \delta ^{\langle 1 \rangle }_x u^{\left( m+1/2 \right) }_{ k } \right) \varDelta x \\&= - \sum _{k=1}^K \frac{1}{2} \delta ^+_x \left( \delta ^-_x u^{\left( m+1/2 \right) }_{ k } \right) ^2 \varDelta x = 0 \end{aligned}$$

proves the latter part. $\square$